check if number belongs to particular base or not

[kamalashraf]

i am making a number base conversion program in c++,
suppose that user enters an base 2 number and want to convert that number in base 4. when he enter the number (which he wants to be converted), how can i detect that number is belongs to base 2 number system or not.
for example: if user enter 1001011001 , program should continue and convert that number to base 4. and if user enter 1200101 or 1001012 , program should not convert that number and output that this number is not belongs to base 2 number system. waiting for your kind reply. thanks.

[keskiverto]

The valid digits for base 2 number are {0, 1}.
The valid digits for base 5 number are {0, 1, 2, 3, 4}.
The valid digits for base 42 number are {0, 1, .. ?}.

You could therefore std::find a character from the string with a comparison functor that returns true, if digit is not valid.

[kamalashraf]

can somebody explain it what he said?

[Bourgond Aries]

Iterate over the number whilst checking if every single char is within the valid range of your operating base.

I think that's what kekiverto meant.

[kamalashraf]

hmmm, i am just new to c++, can you tell me that how can i compare that each digit of a number is within the range?

[Bourgond Aries]

Well first you can create a string object using:

int num = 1001010101;
std::string your_number = std::to_string(int);

Now you can "iterate" over the string, since it works like an array, you can use either [int] or .at(int):

bool is_base_2 = true;
for (int i = 0; i < your_number.size(); ++i)
{
    if (your_number[i] - '0' >= 2)
    {
        is_base_2 = false;
        break;
    }
}
if (is_base_2)
{
    // ... your conversion code
}
else
{
    // ...
}

[cire]

how can i detect that number is belongs to base 2 number system or not

Ask the user. There is no way to determine whether a particular string of digits was meant to be a binary representation or a decimal/octal/hex representation if it consists only of digits which are common to all of them just by inspecting it.